Question

You are constructing a circuit in your predecessor’s notes to try to verify some of their results. The circuit in their notes shows a copper wire with diameter of 0.5 mm, but you only have aluminum that you can draw to any diameter you need (see https://www.youtube.com/watch?v=Q2l6r6751IM). What diameter should you specify to production staff to get the resistance that you need over the same length in the circuit?

Answer 1

The resistivity of the copper wire is

$\rho_{Cu}=1.68\times 10^{-8}\ \Omega.m$

And the resistivity of the aluminum wire is

PAL = 2.65 x 10-8 2.m

The resistance of the copper wire is therefore

$R_{Cu}=\frac{\rho_{Cu}l}{A}$

And the resistance of the aluminum wire with the same length but different cross-section is

$R_{Al}=\frac{\rho_{Al}l}{A'}$

Both the resistance are equal. i.e

$\Rightarrow \frac{\rho_{Cu}l}{A}=\frac{\rho_{Al}l}{A'}$

$\Rightarrow A'=\frac{\rho_{Al}}{\rho_{Cu}}A$

$\Rightarrow \pi (d'/2)^2=\frac{\rho_{Al}}{\rho_{Cu}}\pi (d/2)^2$

$\Rightarrow d'^2=\frac{\rho_{Al}}{\rho_{Cu}} d^2$

$\Rightarrow d'=\sqrt{\frac{\rho_{Al}}{\rho_{Cu}} }d$

Putting the values

$d'=\sqrt{\frac{2.65\times 10^{-8}\ \Omega.m}{1.68\times 10^{-8}\ \Omega.m} }\times 0.5\ mm=0.628\ mm$

So, the diameter of the aluminum wire must be 0.628 mm to get the desired results.